An airplane at T(80,20) needs to fly to both U(20,60) and V(110,85). What is the shortest possible distance for the trip?
(1 point)
Responses
165 units
165 units
170 units
170 units
97 units
97 units
169 units
Using the distance formula, we can find the distance between T and U:
d(TU) = sqrt((20-80)^2 + (60-20)^2) = sqrt(3600 + 1600) = sqrt(5200) ≈ 72.11
Similarly, we can find the distance between T and V:
d(TV) = sqrt((110-80)^2 + (85-20)^2) = sqrt(900 + 4225) = sqrt(5125) ≈ 71.58
The shortest possible distance for the trip would be the smaller of the two distances, which is approximately 71.58 units.
To find the shortest possible distance for the trip, we can use the distance formula. The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the distances between the points:
Distance from T(80,20) to U(20,60):
d1 = sqrt((20 - 80)^2 + (60 - 20)^2) = sqrt((-60)^2 + (40)^2) = sqrt(3600 + 1600) = sqrt(5200) ≈ 72.11 units
Distance from T(80,20) to V(110,85):
d2 = sqrt((110 - 80)^2 + (85 - 20)^2) = sqrt((30)^2 + (65)^2) = sqrt(900 + 4225) = sqrt(5125) ≈ 71.61 units
Therefore, the shortest possible distance for the trip is approximately 71.61 units.